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Lagrangian for a charged particle in a magnetic field

  1. Oct 9, 2007 #1
    Hey I have a question that was almost completely answered in this thread:
    https://www.physicsforums.com/showthread.php?t=116098

    Specifically someone interpreted his question to be how exactly do you get that
    the term in the Lagrangian corresponding to the magnetic potential is

    [tex]-q\vec{v}\cdot\vec{A}[/tex]

    Sadly, I had already worked up to the second to last line prior to his post, and I'm not clear
    on what happened at that point. I would assume he added in [tex]\partial \phi/\partial x[/tex] because the electric potential is zero and that expression will look more like the typical [tex]E_x[/tex] that you see. The part that loses me though is how did he turn the term

    [tex]\dot{x}\frac{\partial A_x}{\partial x}+\dot{y}\frac{\partial A_y}{\partial x}+\dot{z}\frac{\partial A_z}{\partial x}[/tex] into [tex]\dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})[/tex]

    Just from the looks of things it might be that:
    [tex]\dot{x}\frac{\partial A_x}{\partial x} + \dot{y}\frac{\partial A_x}{\partial y} + \dot{z}\frac{\partial A_x}{\partial z} = 0[/tex]

    I don't really see why that should be true, so is there another way he could have done that, or what is the reasoning behind that equation? Also, sorry in advance. I did take a look at Jackson but he jumps into some relativistic stuff that I'm not ready for right now and we don't have a copy of Franklin in our library.
     
  2. jcsd
  3. Oct 9, 2007 #2
    It's explained in the book: "The Classical Theory of Fields" by Landau and Lifschitz.
     
  4. Oct 9, 2007 #3

    clem

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    [tex]
    m\frac{d{\vec v}}{dt}=
    q[-\nabla\phi-\partial_t{\vec A}+{\vec v}\times(\nabla\times{\vec A})]
    = q[-\nabla\phi-\partial_t{\vec A}-({\vec v}\dot\nabla){\vec A}
    +\nabla({\vec v}\cdot{\vec A})]
    =\nabla[-q\phi+q{\vec v}\cdot{\vec A}]-q\frac{d{\vec A}}{dt}.
    [/tex]
     
    Last edited: Oct 9, 2007
  5. Oct 9, 2007 #4
    Alright thanks. It looks like my problem happened here:

    [tex]\frac{d}{dt}(m\dot{x}+qA_x)=q\frac{\partial}{\partial x}(\dot{x}A_x+\dot{y}A_y+\dot{z}A_z)[/tex]

    when I assumed that you could take that derivative of [tex]A_x[/tex] with respect to time partially. If I had taken a total derivative I'd have seen instead:

    [tex]m\ddot{x}+q\frac{\partial A_x}{\partial t} + q(\dot{x}\partial_x A_x + \dot{y}\partial_y A_x + \dot{z} \partial_z A_x) =q(\dot{x}\partial_x A_x +\dot{y}\partial_x A_y+\dot{z}\partial_z A_z)[/tex]

    and that gives me what I was hoping for:

    [tex]m \ddot{x}=-q\frac{\partial A_x}{\partial t}+q[\dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})]=qE_x+q(\dot{y}B_z-\dot{z}B_y)[/tex]
     
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